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Unformatted text preview: Physics 7A Final Exam, Section 1 Prof. Lanzara December 15, 2004 1 Problem 1 (a) Since the system is not accelerating at the beginning of the problem, there is no net force on the system. Therefore, momentum is conserved. mv m + Mv b = 0 v m = M m v b where v m is the man’s speed with respect to the ground and v b is the balloon speed with respect to the ground . We are given the man’s speed with respect to the ladder, which is the difference of v m and v b . Δ v = v m v b = M m v b v b = parenleftbigg 1 + M m parenrightbigg v b 2 km h = parenleftbigg 1 + 400 60 parenrightbigg v b v b = . 261 km h (b) By conservation of momentum, the balloon will be motionless . 1 2 Problem 2 (1) & (2) After the person walks a distance L , the center of mass has moved a distance x cm and the rope has unwound a length l ′ from the spool. We see then that, b b P b x cm l ′ b b l ′ P L L = x cm + l ′ . (1) l ′ is also the distance that the point P has moved on the spool and since the spool rolls without slipping, it is equal to the distance that the center of mass has traveled, x cm . So, using x cm = l ′ we have, l ′ = L 2 . (2) For x cm we can use our result from the first part, x cm = L 2 . (3) We could have also solved this problem by thinking about velocities. Relative to the ground we have for the velocity of the man in terms of the center of mass velocity of the spool and its angular velocity, v man = v cm + Rω (4) The spool rolls without slipping so, v cm = Rω , and we have for the velocity of the man, v man = 2 Rω. (5) 2 We can integrate both sides to relate the angular displacement in the spool (rotating with respect to its center of mass) to the total distance the man travels, L , integraldisplay v man dt = 2 R integraldisplay ωdt L = 2 R Δ θ. R Δ θ is the arclength that the point P travels so it is exactly equal to the length that the rope has been extended. Then, using R Δ θ = l ′ we solve for the rope length, l ′ = L 2 . (6) Since the spool rolls without slipping we have for the center of mass, x cm = R Δ θ (7) = L 2 . (8) (3) Looking at the FBD for the spool, b T F g F N a b h we choose the corner of the step as the pivot and we have for the net torque about that point (choosing clockwise as positive), τ net = ( a + R ) T + bF N bF g . (9) The moment of inertia about the corner is, I = I cm + MR 2 = 1 2 MR 2 + MR 2 = 3 2 MR 2 , 3 so we have for Newton’s second law for a rotating system, τ = Iα → α = (( a + R ) T + bF N bF g ) 3 2 MR 2 ....
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This note was uploaded on 10/17/2011 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at University of California, Berkeley.
 Spring '08
 Lanzara
 Physics, Force, Momentum

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